Question

Which among the following molecule exhibits paramagnetism?

• A. \(\text{O}_2\)
• B. \(\text{O}_3\)
• C. \(\text{N}_2\)
• D. \(\text{F}_2\)

Hint:

A molecule is paramagnetic if it has an unpaired electron in its molecular orbitals; draw the MO configuration.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Paramagnetism comes from unpaired electrons. Molecular orbital (MO) theory shows the electron count in each orbital, so we can see which molecule has an unpaired electron.

Step 2: Fill the MOs for O\(_2\):
O\(_2\) has 16 electrons. The configuration is \(\sigma 1s^2\,\sigma^* 1s^2\,\sigma 2s^2\,\sigma^* 2s^2\,\sigma 2p_z^2\,\pi 2p_x^2 = \pi 2p_y^2\,\pi^* 2p_x^1 = \pi^* 2p_y^1\).
The two electrons in the degenerate \(\pi^*\) orbitals stay unpaired (Hund's rule), so O\(_2\) is paramagnetic.

Step 3: Check the Other Options:
N\(_2\) (14 electrons) fills every orbital up to \(\sigma 2p_z^2\) in pairs, so it is diamagnetic. F\(_2\) (18 electrons) has all the \(\pi^*\) orbitals completely filled, so it is diamagnetic. O\(_3\) has all electrons paired in its resonance structures, so it is diamagnetic.

Step 4: Conclusion:
Only O\(_2\) has unpaired electrons.

Final Answer:
Oxygen is paramagnetic because of two unpaired electrons in \(\pi^*\) orbitals, option (A). \[ \boxed{\text{(A) } \text{O}_2} \]